Polynomial Min Value: Find a & b Real #s

[VertexOperator]

Homework Statement

Find the minimum possible value for a^{2}+b^{2} where a and b are real such that the following equation has at least one real root.

Homework Equations

x^{4}+ax^{3}+bx^{2}+ax+1

The Attempt at a Solution

I tried to find the roots of the equation and then find a relationship between them but that didn't workout very well for me.

 

[mfb]

There are general equations for those roots, but I don't think they are useful here.

- For b=2, there is a nice factorization. This might correspond to the solution, but I don't know how to show that.
- For minimal a^2+b^2, the polynomial has exactly one real root of degree 2. As complex roots come in pairs (for real coefficients), the polynomial can be written as (x+c)^2(x^2+dx+e) for some real c,d,e. It might be possible to solve this directly.

 

[hapefish]

Here is another way to think about it:

If $x^4+ax^3+bx^2+ax+1$ has at least one real root, then that means its graph must intersect the x-axis. Since this is a 4th degree equation with a positive leading coefficient the tails of the graph both go off towards positive infinity. Therefore, if the minimum point on the graph is negative, then the graph crosses the x-axis at some point, and therefore it has real roots.

This would change your problem statement to: "Find the minimum possible value of $a^2+b^2$ such that the minimum value of $x^4+ax^3+bx^2+ax+1$ is less than zero"

This will still be a mess, but perhaps it is a more manageable mess than the original? (I haven't tried to solve it, but if this is a multivariable calculus class perhaps you could turn this into a constrained optimization problem with variables a, b, and x?)

 

[hapefish]

I think this it:

You can show that any real solution of $x^4+ax^3+bx^2+ax+1=0$ is also a solution of $x^4-ax^3+bx^2-ax+1=0$. Then you can combine these two equations and get that this solution must be a solution of $x^4+bx^2+1=0$. But this is great, because you can show from here that $b^2 \geq 4$. Since you know that letting a=0 and b=2 gives you real solution, you can be confident that this also gives you the least solution.

 

[VertexOperator]

How can we show that any real solution of $x^4+ax^3+bx^2+ax+1=0$ is also a solution of $x^4-ax^3+bx^2-ax+1=0$?

 

[epenguin]

That is a 'reciprocal equation' - if x is a root, 1/x is also a root.

It can be solved conveniently dividing it by x² and expressing it in the variable z = (x + 1/x).

You might see a way to answer the question on the way before getting to the solution of the equation.

 

[Bacle2]

I think this it:

You can show that any real solution of $x^4+ax^3+bx^2+ax+1=0$ is also a solution of $x^4-ax^3+bx^2-ax+1=0$. Then you can combine these two equations and get that this solution must be a solution of $x^4+bx^2+1=0$. But this is great, because you can show from here that $b^2 \geq 4$. Since you know that letting a=0 and b=2 gives you real solution, you can be confident that this also gives you the least solution.

Not quite:

$x^4+4x^3+6x^2+4x+1=(x+1)^3$ , but $x^4-4x^3+6x^2-4x+1=(x-1)^3$

(BTW, this gives you a not-so-great lower bound of $4^2+6^2=52$ for $a^2+b^2$ **)

Still, you do know that if there is one real root, there are at least two, since if

you have a real root, your polynomial splits as a product of a monic and a cubic

and every cubic ( or odd-degree poly.) has a real root, since complex roots come in pairs.

So you get $x^4+ax^3+bx^2+ax+1=(x^2+cx+d)(x^2+ex+f)$

where, WOLG the left poly $(x^2+cx+d)$ splits *

You can then play around with the Galois groups to make sure the lefthand poly. splits.

Note too, that if x<0 and b is positive, it makes it harder for your poly. to have a root, so you want it to be small if positve, or,

even better, to have b negative. Similarly, you want |a|>|b| , since , for x negative, the a coefficient lower the value of

f(x) and a raises it.

* If the poly. doesn't split, then you must acquit.

** Note that $x^4+x^3+x+1$ , which is the substitution a=1 and b=0 , has -1 as a double-root.

 

[Dick]

That is a 'reciprocal equation' - if x is a root, 1/x is also a root.

It can be solved conveniently dividing it by x² and expressing it in the variable z = (x + 1/x).

You might see a way to answer the question on the way before getting to the solution of the equation.

Oooh. That's useful. So, in the extreme case where there is exactly one real root, then x=1/x. Nice hint.

 

[Bacle2]

Oooh. That's useful. So, in the extreme case where there is exactly one real root, then x=1/x. Nice hint.

You mean up to multiplicity, right? Otherwise , if there is one real root, there must be at

least two real roots ( up to multiplicity), since then f(x) splits as the product of a monic and

cubic .

And by epenquin's answer , the second poly. is of the form

$x^2+ex+1$ , and the first is $(x±1)^2$

And then it is relatively-clean to expand :

$(x^2+ex+1)(x±1)^2$ and set it equal to f(x) (which I'll do later when I have more time).

 

[VertexOperator]

How is it even possible that there is only 1 real root? Since it is a quartic shouldn't be there 0, 2, or 4?

 

[Bacle2]

How is it even possible that there is only 1 real root? Since it is a quartic shouldn't be there 0, 2, or 4?

Yes, you're right, (I mentioned that in my post); if there is one root, then f(x) is the product

of a monic and a cubic , and the cubic must have at least one real root), so f(x) splits

as the product of 2 monics and a quadratic.

 

[VertexOperator]

(x-1)^4 has one real root. I don't think we are counting multiplicities here.

ok, but I thought we have to count multiplicities?

 

[Bacle2]

But I think it makes a difference to check multiplicities, since there is a difference between a poly. with a double real root and a complex root (with its conjugate) , and having a poly. with four real roots.

Sorry, my posts were all-over the place; this problem is too difficult for me to do it on-and-off while doing something else.

 

[Dick]

ok, but I thought we have to count multiplicities?

You can count them if you want. But having at least one real root doesn't depend on the multiplicity. The idea I'm trying to follow is that as you decrease a^2+b^2 then at some point there is exactly one real root (and yes, it will have multiplicity at least 2 and possibly 4). That leads back to epenguins hint.

 

[Dick]

But I think it makes a difference to check multiplicities, since there is a difference between a poly. with a double real root and a complex root (with its conjugate) , and having a poly. with four real roots.

Sorry, my posts were all-over the place; this problem is too difficult for me to do it on-and-off while doing something else.

That's ok. Now you've got me worried. I'm trying to say that at the minimal value of a^2+b^2 there is exactly one real root. Hence that root satisfies 1/x=x. But suppose at the minimal value there are TWO real roots. Like 1/2 and 2 both multiplicity 2. Then as you decrease a^2+b^2 then both roots go away at the same time. That seems pretty unlikely, but it's still a gap that needs closing.

 

[Bacle2]

That's ok. Now you've got me worried. I'm trying to say that at the minimal value of a^2+b^2 there is exactly one real root. Hence that root satisfies 1/x=x. But suppose at the minimal value there are TWO real roots. Like 1/2 and 2 both multiplicity 2. Then as you decrease a^2+b^2 then both roots go away at the same time. That seems pretty unlikely, but it's still a gap that needs closing.

Sorry, have not been too focused here, but maybe we can use the fact here that

1 is an upper bound for $a^2+b^2$ , since a=1 and b=0 leads to f(x) having -1

as a root.

 

[Dick]

Sorry, have not been too focused here, but maybe we can use the fact here that

1 is an upper bound for $a^2+b^2$ , since a=1 and b=0 leads to f(x) having -1

as a root.

Sure. There's a lot of upper bounds. And 0 is certainly a lower bound where there are no real roots. The trick is to use some property that happens exactly at the lower bound. Keep thinking about it. If I assume there is exactly one real root at the lower bound with multiplicity 2, then I think I've got it. If there are two, I don't. It's a challenging question.

 

[VertexOperator]

This question is from the IMO btw.

 

[mfb]

That is a 'reciprocal equation' - if x is a root, 1/x is also a root.

It can be solved conveniently dividing it by x² and expressing it in the variable z = (x + 1/x).

You might see a way to answer the question on the way before getting to the solution of the equation.

Great. I was sure that the symmetry could be used in some way, but did not see the trick.

This simplifies my equation in post 2...

If I assume there is exactly one real root at the lower bound with multiplicity 2, then I think I've got it. If there are two, I don't. It's a challenging question.

Edit: Hmm, that might be a problem.

 

[voko]

The equation is palindromic, so the standard technique is: $$
x^4 + ax^3 + bx^2 + ax + 1 = 0

\\ \Rightarrow x^2 + ax + b + a/x + 1/x^2 = 0

\\ \Rightarrow X^2 + pX + q = 0 $$ where $X = 1/x + x$. The resultant quadratic is then analyzed.

 

[Dick]

The equation is palindromic, so the standard technique is: $$
x^4 + ax^3 + bx^2 + ax + 1 = 0

\\ \Rightarrow x^2 + ax + b + a/x + 1/x^2 = 0

\\ \Rightarrow X^2 + pX + q = 0 $$ where $X = 1/x + x$. The resultant quadratic is then analyzed.

Yeah, that was epenguin's hint. I was sort of hoping to get away with an argument that didn't require actually finding the roots.

 

[epenguin]

Think about this; you get a quadratic in z .

I think if you think about it you find you can't get a real root in x if z is not real.

So the discriminant (a² - 4b + 8) of the quadratic in z has to be positive. But whatever else that limits, it does not put any minimum on a² nor b² which can be as small as you like. Or rather a can be as small as you like and as long as b is small enough it is OK

So you only have to worry about the equation for x:

x² - xz + 1 = 0

From that, or else just by looking at the function (x + 1/x) you realize that no z between -2 and 2 can correspond to real x. So at least one of the z has to be outside that range. After which it may not be very nice, but you have some handle on it.

Everything I say needs you check.

 

[epenguin]

Yeah, that was epenguin's hint. I was sort of hoping to get away with an argument that didn't require actually finding the roots.

So was I. After the problem has been solved we might find the elegant argument we should have thought of in the first place.

 

[Dick]

So was I. After the problem has been solved we might find the elegant arument we should have thought of in the first place.

Alright then. I think you can handle the two real zero case. If a root r isn't equal to +/-1, then the polynomial having two roots of multiplicity 2 is (x-r)^2*(x-1/r)^2. You can expand it out and find a and b. Then find extrema of a^2+b^2 by differentiating it. It's minimized at r=+/-1, but the value of a^2+b^2 is 52. Which is way too large, since Bacle found a case where it's 1. The other possibility is (x-1)^2*(x+1)^2. Which gives a^2+b^2=4. Still too big.

That blemish eliminated then we should be back on track. In the case of minimal a^2+b^2 there is one real root of multiplicity 2. Hence its value must be +1 or -1. The rest is easy.

 

[epenguin]

Oof, just as I predicted, having convinced myself by the argument I indicated of what the answer was, I looked at the equation and it is indeed obvious that the minimum (a² + b²) for real roots is 0, since in that case it has 1 and -1 as roots.

No doubt the smartasses who compete in this despicable IMO competion are wised up to look for that sort of rubbish.

 

[Dick]

Oof, just as I predicted, having convinced myself by the argument I indicated of what the answer was, I looked at the equation and it is indeed obvious that the minimum (a² + b²) for real roots is 0, since in that case it has 1 and -1 as roots.

No doubt the smartasses who compete in this despicable competion are wised up to look for that sort of rubbish.

Mmm. x^4+1=0 doesn't have any real roots. What do you mean? There IS a positive value of a^2+b^2 where there are real solutions and below which there aren't. I don't usually care much for Olympiad problems because they require more brains than I'm usually willing to expend (or have) on getting some small result. Wouldn't call them despicable or smartasses though. Some people like this stuff. And it actually was kind of entertaining. Kept me off the streets for a while.

 

[epenguin]

Mmm. x^4+1=0 doesn't have any real roots. What do you mean? There IS a positive value of a^2+b^2 where there are real solutions and below which there aren't. I don't usually care much for Olympiad problems because they require more brains than I'm usually willing to expend (or have) on getting some small result. Wouldn't call them despicable or smartasses though. Some people like this stuff. And it actually was kind of entertaining. Kept me off the streets for a while.

OK it was late at night. However it does maybe suggest an approach to the answer.

 

[Dick]

OK it was late at night. However it does maybe suggest an approach to the answer.

The solution I did (thanks to your hint) is "almost" really easy. If there is a single real root of multiplicity 2 then it must be 1 or -1. Then you just have to extremize a^2+b^2 subject to the linear constraint you get from putting 1 or -1 in as a root. That's easy. The glitch I ran into was excluding the case that at the minimal value of a^2+b^2 there might be two real roots of multiplicity 2. That made things a little ugly. But doable.

 

[VertexOperator]

So what is the final answer to this question :(

 

[Dick]

So what is the final answer to this question :(

Shouldn't you try to work it out for yourself? There's hints galore here.